MathsCBSE Class 10 · Coordinate Geometry

Midpoint Formula and Points of Trisection, Class 10

Updated 13 October 2026 · Fact-checked

The midpoint of a segment is the point exactly halfway between its ends, found by averaging the x-coordinates and the y-coordinates. Points of trisection are the two points that cut a segment into three equal parts. The board asks you to find these points, find a missing end or unknown, and check positions on the axes.

One third of the total step (2, 1) is added once to reach P, twice to reach Q and three times to reach B, while M sits halfway between P and Q.

Illustration drawn for QuestionBuddy.

Understand Midpoint Formula and Points of Trisection

The midpoint of a segment AB is the point M halfway between A and B. It is the special case of the section formula where the ratio is $1 : 1$ . Put $m_{1} = m_{2} = 1$ in the section formula and you get $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ . In words, each coordinate of the midpoint is the average of the two end coordinates.

Why does averaging work? Going from A to M you move half of the total horizontal step and half of the total vertical step. So $x_{M} = x_{1} + \frac{x _{2} - x _{1}}{2} = \frac{x _{1} + x _{2}}{2}$ , and the same for y. This is the same similar-triangles idea that proves the section formula.

Points of trisection split AB into three equal parts, so there are always two of them. P is one third of the way from A, so $A P : P B = 1 : 2$ . Q is two thirds of the way, so $A Q : QB = 2 : 1$ . You find each with the section formula. A faster way is the step method: the total step from A to B is $(x_{2} - x_{1}, y_{2} - y_{1})$ . One third of it is the step for each equal part. Add it to A once to get P, twice to get Q, and three times to get B, which is a free check.

The same step idea works for any number of equal parts. To cut AB into four equal parts, use a quarter step. The three points are at ratios $1 : 3$ , $1 : 1$ and $3 : 1$ from A, and the middle one is the midpoint.

The formulas also run backwards. If you know the midpoint M and one end A, the other end is $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$ . If you know both trisection points, their difference is one step, so you can rebuild A and B.

These results need the points to be on the segment, which is internal division. The midpoint and trisection points always lie between A and B, so their coordinates lie between the end coordinates. Use this as a quick sense check every time.

Midpoint Formula and Points of Trisection: formulas to remember

Midpoint formula: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$; Midpoint of $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ . Section formula with ratio $1 : 1$ .
Missing end from the midpoint: $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$; Use when M is the midpoint of AB and A is known.
First point of trisection (ratio 1 : 2): $P = (\frac{x _{2} + 2 x _{1}}{3}, \frac{y _{2} + 2 y _{1}}{3})$; P is nearer A. Note that the weight 2 goes with A.
Second point of trisection (ratio 2 : 1): $Q = (\frac{2 x _{2} + x _{1}}{3}, \frac{2 y _{2} + y _{1}}{3})$; Q is nearer B. Q is also the midpoint of PB.
Step method for n equal parts: $step = (\frac{x _{2} - x _{1}}{n}, \frac{y _{2} - y _{1}}{n})$; Add the step to A repeatedly. Use n = 3 for trisection and n = 4 for four equal parts.
Points dividing AB into four equal parts: $ratios 1 : 3, 1 : 1, 3 : 1 from A$; The middle point is the midpoint of AB.

How to solve Midpoint Formula and Points of Trisection questions

This method works for every midpoint, trisection and equal-parts question, including ones with unknowns.

1Write the end points clearly as $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ , and note how many equal parts are needed.
2For a midpoint, average the x-coordinates and the y-coordinates separately.
3For trisection, find P with ratio $1 : 2$ and Q with ratio $2 : 1$ using the section formula, pairing $m_{1}$ with B.
4Check with the step method: one step is $\frac{1}{3}$ of $(x_{2} - x_{1}, y_{2} - y_{1})$ ; A plus one step gives P, plus two steps gives Q.
5If the point is given and an end or a letter is unknown, equate coordinates and solve each linear equation separately.
6Sense check: every answer must lie between A and B. State both trisection points with units if a context is given.

Midpoint Formula and Points of Trisection in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Midpoint Formula and Points of Trisection

Suppose your housing society strings festival lights on a straight wire between two poles. On a plan marked in metres, pole A is at $(- 4, 2)$ and pole B is at $(8, 11)$ . The organisers want two lanterns that cut the wire into three equal lengths and a big star exactly at the middle of the wire.

Find the positions of the two lanterns and the star. Also find the distance between the two lanterns and the total length of the wire.

Try it first, then show the solutionSolution

Total step from A to B: $(8 - (- 4), 11 - 2) = (12, 9)$ .
One third of the step: $(4, 3)$ .
First lantern: $(- 4 + 4, 2 + 3) = (0, 5)$ .
Second lantern: $(0 + 4, 5 + 3) = (4, 8)$ ; check $(4 + 4, 8 + 3) = (8, 11) = B$ .
Star at midpoint: $(\frac{- 4 + 8}{2}, \frac{2 + 11}{2}) = (2, 6.5)$ .
Distance between lanterns: $4^{2} + 3^{2} = 25 = 5$ m.
Wire length: $1 2^{2} + 9^{2} = 225 = 15$ m, which is $3 \times 5$ m as expected.

Answer: Lanterns at $(0, 5)$ and $(4, 8)$ , star at $(2, 6.5)$ ; lanterns are 5 m apart and the wire is 15 m long.

Memory hook: Equal lanterns, equal steps: trisection is just one third of the journey added again and again.

Tricks to remember Midpoint Formula and Points of Trisection

Short memory aids. Say each one aloud once, then use it on a question the same day.

Midpoint = average
Add and halve: the midpoint is the average of the ends, coordinate by coordinate.
Use it when: Every midpoint and parallelogram-diagonal question.
Double minus for the missing end
Other end = twice the middle minus the known end.
Use it when: When the midpoint and one end are given.
Two trisection points, near weight goes far
For the point nearer A, the heavier weight 2 sits on A: $\frac{x _{2} + 2 x _{1}}{3}$ . The bigger weight pulls the point towards its own end.
Use it when: Avoid swapping the ratio when writing P and Q.
Step, step, step
One third of $B - A$ is one step. A + 1 step = P, + 2 steps = Q, + 3 steps = B (the check).
Use it when: Fast working and verification of trisection points.

Midpoint Formula and Points of Trisection: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Unknown end from a midpoint · 1 mark · MCQBoard level

The midpoint of the segment joining $A (- 4, 7)$ and $B (x, - 3)$ is $(2, y)$ . The value of $x + y$ is:
(a) 6
(b) 10
(c) 8
(d) 12

Try it first, then show the solutionSolution

x-coordinate: $\frac{- 4 + x}{2} = 2$ , so $x = 8$ .
y-coordinate: $y = \frac{7 + ( - 3 )}{2} = 2$ .
$x + y = 8 + 2 = 10$ .

Answer: (b) 10

Example 2: Points of trisection with a check · 3 marks · Short answerBoard level

Find the coordinates of the points of trisection of the segment joining $A (2, - 2)$ and $B (- 7, 4)$ . Show that one of them lies on the x-axis.

Try it first, then show the solutionSolution

Let P and Q trisect AB with $A P : P B = 1 : 2$ and $A Q : QB = 2 : 1$ . (½ mark)
$P = (\frac{1 ( - 7 ) + 2 ( 2 )}{3}, \frac{1 ( 4 ) + 2 ( - 2 )}{3}) = (- 1, 0)$ . (1 mark)
$Q = (\frac{2 ( - 7 ) + 1 ( 2 )}{3}, \frac{2 ( 4 ) + 1 ( - 2 )}{3}) = (- 4, 2)$ . (1 mark)
P has y-coordinate 0, so P lies on the x-axis. Check: Q is the midpoint of PB, $(\frac{- 1 - 7}{2}, \frac{0 + 4}{2}) = (- 4, 2)$ . (½ mark)

Answer: $P (- 1, 0)$ and $Q (- 4, 2)$ ; P lies on the x-axis.

Example 3: Rebuilding the segment from its trisection points · 3 marks · Short answerHarder board level

The points $P (- 1, 3)$ and $Q (2, 5)$ trisect the segment AB, with P nearer A. Find the coordinates of A and B.

Try it first, then show the solutionSolution

Since $A P = P Q = QB$ , P is the midpoint of AQ and Q is the midpoint of PB. (½ mark)
From P midpoint of AQ: $A = (2 (- 1) - 2, 2 (3) - 5) = (- 4, 1)$ . (1 mark)
From Q midpoint of PB: $B = (2 (2) - (- 1), 2 (5) - 3) = (5, 7)$ . (1 mark)
Check: step $Q - P = (3, 2)$ ; $A + (3, 2) = (- 1, 3) = P$ and $Q + (3, 2) = (5, 7) = B$ . (½ mark)

Answer: $A (- 4, 1)$ and $B (5, 7)$

Example 4: Dividing a segment into four equal parts · 5 marks · Long answerHarder board level

Find the coordinates of the points which divide the segment joining $A (- 2, 3)$ and $B (6, - 5)$ into four equal parts. Show that the first of these points lies on the y-axis, find the length of each part, and verify that four parts make up AB.

Try it first, then show the solutionSolution

Let $P_{1}, P_{2}, P_{3}$ divide AB into four equal parts, at ratios $1 : 3$ , $1 : 1$ , $3 : 1$ from A. (½ mark)
$P_{1} = (\frac{1 ( 6 ) + 3 ( - 2 )}{4}, \frac{1 ( - 5 ) + 3 ( 3 )}{4}) = (0, 1)$ . (1 mark)
$P_{2}$ is the midpoint: $(\frac{- 2 + 6}{2}, \frac{3 - 5}{2}) = (2, - 1)$ . (1 mark)
$P_{3} = (\frac{3 ( 6 ) + 1 ( - 2 )}{4}, \frac{3 ( - 5 ) + 1 ( 3 )}{4}) = (4, - 3)$ . (1 mark)
$P_{1}$ has x-coordinate 0, so it lies on the y-axis. (½ mark)
Each part: $A P_{1} = (0 + 2)^{2} + (1 - 3)^{2} = 8 = 22$ units. (½ mark)
$A B = 8^{2} + 8^{2} = 82 = 4 \times 22$ units, verified. (½ mark)

Answer: $(0, 1)$ , $(2, - 1)$ , $(4, - 3)$ ; each part is $22$ units and $A B = 82$ units.

Example 5: Case study: sports day track · 4 marks · Case-basedHOTS / competency

Imagine your school marks a straight running track on the ground, on a grid in metres, from start S at $(- 3, 8)$ to finish F at $(9, - 4)$ . Two water stations $W_{1}$ and $W_{2}$ divide SF into three equal parts, with $W_{1}$ nearer S. A first-aid tent is placed at $T (3, 2)$ .
(i) Find the position of $W_{1}$ . (1 mark)
(ii) The coach claims the tent is exactly midway between the two water stations. Find $W_{2}$ and decide, with a reason, whether the coach is right. (1 mark)
(iii) A camera is to be fixed on the x-axis so that it is equally far from both water stations. Find its position and its distance from each station. (2 marks)

Try it first, then show the solutionSolution

(i) $W_{1}$ divides SF in $1 : 2$ : $W_{1} = (\frac{9 + 2 ( - 3 )}{3}, \frac{- 4 + 2 ( 8 )}{3}) = (1, 4)$ . (1 mark)
(ii) $W_{2}$ divides SF in $2 : 1$ : $W_{2} = (\frac{2 ( 9 ) + ( - 3 )}{3}, \frac{2 ( - 4 ) + 8}{3}) = (5, 0)$ . (½ mark)
Midpoint of $W_{1} W_{2}$ : $(\frac{1 + 5}{2}, \frac{4 + 0}{2}) = (3, 2) = T$ , so the coach is right. This is expected, since the two stations sit equally far from the middle of SF. (½ mark)
(iii) Let the camera be $C (x, 0)$ . $C W_{1} = C W_{2}$ gives $(x - 1)^{2} + (0 - 4)^{2} = (x - 5)^{2} + (0 - 0)^{2}$ . (½ mark)
$x^{2} - 2 x + 17 = x^{2} - 10 x + 25$ , so $8 x = 8$ . (½ mark)
$x = 1$ , so the camera is at $C (1, 0)$ . (½ mark)
Check: $C W_{1} = 0^{2} + 4^{2} = 4$ m and $C W_{2} = 4^{2} + 0^{2} = 4$ m, equal as required. (½ mark)

Answer: (i) $W_{1} (1, 4)$ (ii) $W_{2} (5, 0)$ ; the midpoint of $W_{1} W_{2}$ is $(3, 2)$ , so the coach is right. (iii) Camera at $(1, 0)$ , 4 m from each station.

Example 6: Fourth vertex of a parallelogram with a side check · 3 marks · Short answerHarder board level

The points $A (1, 2)$ , $B (4, y)$ , $C (x, 6)$ and $D (3, 5)$ , taken in order, are the vertices of a parallelogram. Find x and y. Then show that ABCD is not a rhombus.

Try it first, then show the solutionSolution

The diagonals of a parallelogram bisect each other, so the midpoint of AC equals the midpoint of BD. (½ mark)
x-coordinates: $\frac{1 + x}{2} = \frac{4 + 3}{2}$ , so $x = 6$ . (½ mark)
y-coordinates: $\frac{2 + 6}{2} = \frac{y + 5}{2}$ , so $y = 3$ . (½ mark)
$A B = (4 - 1)^{2} + (3 - 2)^{2} = 10$ units and $B C = (6 - 4)^{2} + (6 - 3)^{2} = 13$ units. (1 mark)
Adjacent sides AB and BC are not equal, so ABCD is not a rhombus. (½ mark)

Answer: $x = 6$ , $y = 3$ ; since $A B = 10 \neq = B C = 13$ , ABCD is not a rhombus.

Example 7: Assertion-reason on the midpoint · 1 mark · Assertion-reasonBoard level

Assertion (A): The midpoint of the segment joining $(2, 3)$ and $(4, 7)$ is $(3, 5)$ .
Reason (R): The midpoint of a segment is the point given by the section formula with ratio $1 : 1$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

With $m_{1} = m_{2} = 1$ , the section formula gives $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ , so R is true.
Using it: $(\frac{2 + 4}{2}, \frac{3 + 7}{2}) = (3, 5)$ , so A is true.
A is obtained directly by applying R, so R correctly explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Where students lose marks in Midpoint Formula and Points of Trisection

The slips examiners see most often, and what to write instead.

Mistake: Giving only one point of trisection
Students stop after the first point, forgetting a segment has two trisection points.
Wrong working
Points of trisection of $A (2, - 2)$ , $B (- 7, 4)$ : $(- 1, 0)$ only.
Right working
$P (- 1, 0)$ with ratio $1 : 2$ and $Q (- 4, 2)$ with ratio $2 : 1$ .
Do this: Always write 'Let P and Q be the points of trisection' and find both.
Usually costs: About half the marks of the question.
Mistake: Putting the weight 2 on the wrong end
The cross pairing of the section formula is forgotten, so the point nearer A is found with weight 2 on B.
Wrong working
Point nearer A for $A (2, - 2)$ , $B (- 7, 4)$ : $(\frac{2 ( - 7 ) + 2}{3}, \frac{2 ( 4 ) - 2}{3}) = (- 4, 2)$ , which is actually nearer B.
Right working
$(\frac{- 7 + 2 ( 2 )}{3}, \frac{4 + 2 ( - 2 )}{3}) = (- 1, 0)$ .
Do this: The point nearer A gets the bigger weight on A. Check with the step method.
Usually costs: 1 mark or more, and wrong follow-on answers.
Mistake: Subtracting instead of adding in the midpoint formula
The distance formula uses differences, and students mix the two formulas.
Wrong working
Midpoint of $(- 6, 4)$ , $(2, - 8)$ : $(\frac{- 6 - 2}{2}, \frac{4 + 8}{2}) = (- 4, 6)$ .
Right working
$(\frac{- 6 + 2}{2}, \frac{4 + ( - 8 )}{2}) = (- 2, - 2)$ .
Do this: Distance subtracts, midpoint adds. Check the answer lies between the ends.
Usually costs: The full mark in an MCQ, and step marks later.
Mistake: Averaging the midpoint and the known end to find the missing end
Students apply the midpoint formula to the wrong pair of points.
Wrong working
M(1, 3) is the midpoint of AB with A(5, 7): B = $(\frac{5 + 1}{2}, \frac{7 + 3}{2}) = (3, 5)$ .
Right working
$B = (2 (1) - 5, 2 (3) - 7) = (- 3, - 1)$ ; check midpoint of $(5, 7)$ and $(- 3, - 1)$ is $(1, 3)$ .
Do this: Set $\frac{5 + x}{2} = 1$ and $\frac{7 + y}{2} = 3$ , or use 'twice the middle minus the known end'.
Usually costs: Usually the whole 2-mark question.

Exam tips

Write the ratio before the formula, for example 'P divides AB in $1 : 2$ '. Examiners give a step mark for it.
For trisection, show both points and add a one-line check (step method or Q is the midpoint of PB).
When a question says a point lies on an axis, set the other coordinate to zero and say so in words.
In case studies, read which end is the start, because 'the station nearer S' fixes which ratio to use. Give units in every sub-part.
Leave coordinates as exact fractions or decimals like $(2, 6.5)$ rather than rounding.

Practice questions on Midpoint Formula and Points of Trisection

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
The point $(3, - 1)$ is the midpoint of the segment joining $A (2 a + 1, 5)$ and $B (7, b - 3)$ . Find a and b.
Hint
Equate the x-coordinates and the y-coordinates separately.
Answer
$\frac{2 a + 8}{2} = 3$ gives $a = - 1$ ; $\frac{b + 2}{2} = - 1$ gives $b = - 4$ .
Q2 · 3 marks · Short answerHarder board level
A is $(k, 5)$ and B is $(8, - 1)$ . The point of trisection of AB nearer A lies on the line $x - y = 1$ . Find k and both points of trisection.
Hint
Write the point nearer A in terms of k first.
Answer
Point nearer A (ratio $1 : 2$ ): $(\frac{8 + 2 k}{3}, \frac{- 1 + 10}{3}) = (\frac{8 + 2 k}{3}, 3)$ . On $x - y = 1$ : $\frac{8 + 2 k}{3} = 4$ , so $k = 2$ . Points of trisection: $(4, 3)$ and $(\frac{16 + 2}{3}, \frac{- 2 + 5}{3}) = (6, 1)$ . Check: step $(2, - 2)$ takes $(2, 5)$ to $(4, 3)$ , $(6, 1)$ , $(8, - 1)$ .
Q3 · 3 marks · Short answerHarder board level
A is $(- 1, 4)$ and B is $(8, y)$ . The point of trisection of AB nearer A is $(2, 6)$ . Find y and the other point of trisection.
Hint
Use ratio $1 : 2$ for the y-coordinate, then ratio $2 : 1$ for the other point.
Answer
$\frac{y + 2 ( 4 )}{3} = 6$ gives $y = 10$ . Other point: $(\frac{16 - 1}{3}, \frac{20 + 4}{3}) = (5, 8)$ .
Q4 · 5 marks · Long answerHarder board level
P and Q trisect the segment joining $A (4, - 6)$ and $B (- 2, 6)$ , with P nearer A. (i) Find P and Q. (ii) Show that Q lies on the y-axis. (iii) If P lies on the line $3 x + k y = 2$ , find k. (iv) Find PQ and verify that $A B = 3 \times P Q$ .
Hint
Step $= \frac{1}{3} (- 6, 12)$ .
Answer
Step $(- 2, 4)$ : $P (2, - 2)$ , $Q (0, 2)$ . Q has x = 0, so it is on the y-axis. $6 - 2 k = 2$ gives $k = 2$ . $P Q = 4 + 16 = 25$ units; $A B = 36 + 144 = 65 = 3 \times 25$ units.
Q5 · 2 marks · Short answerHOTS / competency
A student found the midpoint of $(- 6, 4)$ and $(2, - 8)$ as $(\frac{- 6 - 2}{2}, \frac{4 + 8}{2}) = (- 4, 6)$ . Explain the error and find the correct midpoint. Give a quick check that shows the student's answer cannot be right.
Hint
The midpoint must lie between the two ends.
Answer
The student subtracted the coordinates instead of adding them. Correct midpoint: $(\frac{- 6 + 2}{2}, \frac{4 - 8}{2}) = (- 2, - 2)$ . Check: the y-coordinate 6 is not between 4 and $- 8$ , so $(- 4, 6)$ cannot lie on the segment.
Q6 · 4 marks · Case-basedHOTS / competency
Suppose a circular flower bed is drawn on a park plan marked in metres. A straight path runs across the bed along a diameter from $A (- 3, 2)$ to $B (5, 8)$ . (i) A sprinkler is to be placed at the centre. Find its position. (1 mark) (ii) Find the radius of the bed. (1 mark) (iii) A lamp post stands at $P (4, 1)$ and a bench at $Q (6, 4)$ . Decide for each whether it is on the edge of the bed, inside it or outside it, and justify. (2 marks)
Hint
The centre is the midpoint of a diameter. Compare each distance from the centre with the radius.
Answer
(i) Centre $O = (\frac{- 3 + 5}{2}, \frac{2 + 8}{2}) = (1, 5)$ . (ii) $O A = (1 + 3)^{2} + (5 - 2)^{2} = 25 = 5$ m. (iii) $O P = 3^{2} + (- 4)^{2} = 5$ m, equal to the radius, so the lamp post is on the edge. $O Q = 5^{2} + (- 1)^{2} = 26$ m, which is more than 5 m, so the bench is outside the bed.
Q7 · 3 marks · Short answerHarder board level
P is the point of trisection of the segment joining $A (- 1, 7)$ and $B (5, 1)$ that is nearer A, and M is the midpoint of AB. Find the point on the y-axis that is equidistant from P and M, and verify your answer.
Hint
Find P and M first, then take the point as $(0, y)$ and equate squared distances.
Answer
$P = (\frac{5 + 2 ( - 1 )}{3}, \frac{1 + 2 ( 7 )}{3}) = (1, 5)$ and $M = (\frac{- 1 + 5}{2}, \frac{7 + 1}{2}) = (2, 4)$ . Let $R (0, y)$ : $1 + (y - 5)^{2} = 4 + (y - 4)^{2}$ gives $26 - 10 y = 20 - 8 y$ , so $y = 3$ . $R (0, 3)$ ; check $R P = 1 + 4 = 5$ units and $R M = 4 + 1 = 5$ units.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Midpoint of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ?Show answer
$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
2.Ratios for the two points of trisection from A?Show answer
$1 : 2$ and $2 : 1$
3.Missing end B if M is the midpoint of AB?Show answer
$B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$
4.Midpoint of $(- 3, 5)$ and $(7, - 1)$ ?Show answer
$(2, 2)$
5.Ratios for the points dividing AB into four equal parts?Show answer
$1 : 3$ , $1 : 1$ , $3 : 1$ from A
6.Which trisection point is the midpoint of PB?Show answer
Q, the point nearer B

Midpoint Formula and Points of Trisection: frequently asked questions

Is the midpoint formula a separate formula or part of the section formula?

It is the section formula with the ratio 1 : 1. Because both weights are equal, the formula reduces to the average of the two ends. You can quote it directly in the board exam.

How many points of trisection does a line segment have?

Exactly two. One divides the segment in the ratio 1 : 2 and the other in the ratio 2 : 1, measured from the first end. Always give both.

Can I use the step method in the board exam instead of the section formula?

Yes, any correct method earns marks if your working is clear. Many students use the section formula for the answer and the step method as a check, which is a safe habit.

Is external division or area of a triangle from coordinates in the syllabus?

No. Both were removed and are not in the 2026-27 syllabus, so only internal division is examined.